Code:

char *buf;

This doesn't create memory for you. All a pointer is is an integer describing where memory is. And just like an uninitialized integer, an uninitialized pointer can be anything at all. So you could be trying to use any memory at all, which, in DOS, could mean accidentally overwriting bits of your operating system and hanging the computer or worse... In Linux though, your process resides in its own protected memory space, and if you try to use memory you didn't ask for, it can tell the difference and kills it.

IOW, this code was always wrong. It didn't crash in DOS mode only because DOS mode is physically incapable of detecting that problem. If you compiled this as a 32-bit Windows program it'd crash too.

If you want something that actually has memory, you can declare it as a buffer on the stack:

Code:

char buf[20];

or you can give the pointer something to point to like

Code:

char *buf=malloc(20);

...
// before main returns.  Especially important in DOS!
free(buf);

malloc() needs malloc.h in Turbo C++, or stdlib.h for compilers that aren't 30 years out of date.

The rest of your code looks fine. You should check if fp and fp2 are NULL though -- if they didn't succeed in opening for some reason, your program goes ahead and tries to use them anyway, which will crash too -- or at least really ought to crash.

It might be better to program in Linux than DOS. Linux is a less forgiving environment, you'll catch some mistakes immediately which in DOS might not do anything immediately but could cause very weird side-effects later. Linux is even capable of detecting which line it crashed in if you compile with -ggdb and run the program with gdb...